[Math] Extra solutions when solving $\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$

trigonometry

The problem:

$$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$$

My solution:

$$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$$

$$\sin\theta+\cos\theta=\sqrt{2\times2\sin\theta\cos\theta}$$

$$\sin\theta+\cos\theta=2\sqrt{\sin\theta\cos\theta}$$

$$\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta=0$$

$$(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2=0$$

$$\sqrt{\sin\theta}=\sqrt{\cos\theta}$$

Either $\cos\theta$ or $\sin\theta$ can be equal to zero. Both can't be equal to zero at the same time. As we can see that $\sqrt{\sin\theta}=\sqrt{\cos\theta}$, neither of $\cos\theta$ and $\sin\theta$ is equal to zero. So, dividing both sides by $\cos\theta$ is valid.

$$\sqrt{\tan\theta}=1$$

$$\tan\theta=1$$

$$\theta=n\pi+\frac{\pi}{4}\tag{1}$$

My question:

  1. We've stumbled upon an interesting solution in (1). Here, $\theta$ satisfies our original equation only when $n$ is even or $0$. Why is that? Isn't $n$ supposed to belong to the set of integers?

PS: This might help you in answering the question.

Best Answer

My solution: $$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$$

$$\sin\theta+\cos\theta=\sqrt{2\times2\sin\theta\cos\theta}$$

$$\sin\theta+\cos\theta=2\sqrt{\sin\theta\cos\theta}$$

$$\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta=0$$

$$(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2=0\tag{*}$$

$$\sqrt{\sin\theta}=\sqrt{\cos\theta}$$

$$\sqrt{\tan\theta}=1\tag#$$

$$\tan\theta=1$$

$$\theta=n\pi+\frac{\pi}{4}$$

  1. Note that $$\frac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}=1\implies\sqrt{\frac{\sin\theta}{\cos\theta}}=1$$ but $$\sqrt{\frac{\sin\theta}{\cos\theta}}=1\kern.6em\not\kern -.6em \implies\frac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}=1;$$ so, $$\dfrac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}\not\equiv\sqrt{\tan\theta}.$$ In step $(\#)$, you introduced extraneous solutions—expanding the set of candidate solutions—by allowing $\sin\theta$ and $\cos\theta$ to be both negative (as well as both positive).

    Nevertheless, $(\#)$ is a valid step, as the forward implication is correct.

  2. However, the above may be moot, since the prior step $(*)$ may have discarded solutions and thus may be invalid: without justification, it is not apparent that $$\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta=0\implies(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2=0,$$ since \begin{align}&{}{}\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta\\\not\equiv&{}{}\sin\theta-2\sqrt{\sin\theta}\sqrt{\cos\theta}+\cos\theta\\\equiv&{}{}(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2.\end{align}

    (The converse is certainly true though.)

    Here, the forward-implication does turn out to be justifiable: $\theta$ happens to reside in the first quadrant, so $\sin\theta$ and $\cos\theta$ are indeed both positive.

Addendum at OP's request: \begin{align}&{}\sin\theta+\cos\theta=\sqrt{2\sin2\theta}\\ \color{red}\implies &{}1+\sin2\theta=2\sin2\theta\\ \iff &{}\sin2\theta=1\\ \iff&{}\theta=(4n+1)\frac\pi4.\tag1\end{align} The first step above turns out to have introduced extraneous solutions $(\text{for example, }\frac{5\pi}4),$ which we must prune out. Alternatively, we could note the implicit condition (doing so changes the $\color{red}\implies$ to $\color{red}\iff)$ that $$\sin\theta+\cos\theta\geq0\\ \sqrt2\sin\left(\theta+\frac\pi4\right)\geq0\\ \theta+\frac\pi4\in\bigcup[2n\pi,(2n+1)\pi]$$ $$\theta\in\bigcup[(8n-1)\frac\pi4,(8n+3)\frac\pi4]\tag2.$$ Combining $(1)$ and $(2)$: $$\theta=(8k+1)\frac\pi4\\=\frac\pi4+2k\pi.$$

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